How to find the following integration

Question

Let $X_1, \cdots, X_n$ be $iid$ normal random variables with unknown mean $\mu$ and known variance $\sigma^2$. How to find $E[\Phi(\bar X)]$, where $\bar X:=\frac{\sum_{i=1}^nX_i}{n}$, please? I guess the answer should be $\Phi(\mu)$. Here is how I started. Note that $Y:= \bar X$ is also normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$.

$$E[\Phi(\bar X)] = \int_{-\infty}^\infty \Phi(y) f_Y(y) dy.$$

Let $z=\frac{y-\mu}{\sigma/\sqrt{n}}$ and the above integration becomes

$$\int_{-\infty}^\infty \Phi(\mu+\frac{\sigma z}{\sqrt{n}}) \phi(z) dz$$

I could not proceed any further from here. Does anyone know what to do, please? Thank you!

Answer 1

I don't know how you got to the displayed integral in your question, but here is how to evaluate it once you get there. The result does not match what you claim is the answer.

Let $Y = \Phi(X)$ where $X \sim N(\mu, \sigma^2)$. Then, $$\begin{align} E[Y] = E[\Phi(X)] &= \int_{-\infty}^\infty \Phi(x)f_X(x)\,\mathrm dx \tag{1} \end{align}$$ But, if $Z\sim N(0,1)$ is independent of $X$, then $$P\{Z \leq x\mid X = x\} = P\{Z \leq x\} = \Phi(x)$$ and so we can recognize the integral in $(1)$ as $$\begin{align} E[\Phi(X)] &= \int_{-\infty}^\infty \Phi(x)f_X(x)\,\mathrm dx\\ &= \int_{-\infty}^\infty P\{Z \leq x\mid X = x\}f_X(x)\,\mathrm dx\\ &= P\{Z \leq X\}&{\scriptstyle{\text{by the continuous version of the law of total probability}}}\\ &= P\{Z-X \leq 0\}\\ &= \Phi\left(\frac{0-(-\mu)}{\sqrt{1 + \sigma^2}}\right) &{\scriptstyle{\text{this follows because}~ Z-X \sim N(-\mu,1+\sigma^2)}}\\ &= \Phi\left(\frac{\mu}{\sqrt{1 + \sigma^2}}\right) \end{align}$$

Answer 2

Find $E[\Phi(c\bar X)]$ where $X_i \sim N(\mu, 1)$ and $c$ is a constant. Then, proceed following Dilip but replace $X$ with $\bar Xc.$ Let $Z\sim N(0,1)$ be independent of all $X_i.$ Then $$\begin{align} E[\Phi(c\bar X)] &= \int_{-\infty}^\infty P\{Z \leq cy\mid \bar X = y\}f_\bar X(y)\,\mathrm dy\\ &= P\{Z \leq c\bar X\}\\ &= P\{Z-c\bar X \leq 0\} \\ &= \Phi\left(\frac{c\mu}{\sqrt{1 + c^2/n}}\right) \end{align}$$ If we then choose $c$ to make this last expression equal to $\Phi(\mu),$ then we will obtain its MVUE. In this case $c=\sqrt\frac n{n-1}.$ If $\sigma^2\ne1$, then $c=\sqrt\frac n{n-\sigma^2}$ for $n$ sufficiently large.